18 
Vol.12 No.2 (June 2011) 

                                                          (Short communications)        

Iraqi Journal of Chemical and Petroleum Engineering 
 Vol.12 No.2 (June 2011) 18 – 20  

ISSN: 1997-4884 

 

 

A PARTICULAR SOLUTION OF THE TWO AND THREE                                                      

DIMENSIONAL TRANSIENT DIFFUSION EQUATIONS 

                                            Adel Al-Hemiri  

                 Chemical Engineering Department – University of Baghdad 

 

ABSTRACT 

      A particular solution of the two and three dimensional unsteady state thermal or 

mass diffusion equation is obtained by introducing a combination of variables of the 

form,  

η = (x+y) / √ct , and  η = (x+y+z) / √ct, for two and three dimensional equations  

respectively. And the corresponding solutions are, 

 θ t, x, y =  θ0   erfc 
x +y

√8ct
         and        θ t, x, y, z = θ0 erfc   

x +y +z

√12ct
    

 

Keywords:  Two and three dimensional equations, Particular solution. 

 

INTRODUCTION 

     The unsteady state two and three 

dimensional diffusion equations may be 

solved by the method of Fourier transform 

(separation of variables) or by numerical 

methods to obtain a general solution.  

However both these methods leads to a 

double or triple series of the characteristic 

function and two or three separate 

expansion  problems- depending, of course, 

on whether the equation is two or three 

dimensional- which is a difficult task that 

requires rigorous calculations[1,2,3]. Thus 

in this work we present a solution easily 

obtained  using the method of   combination  

 

of variables in the same manner it was used 

to solve the one dimensional equation: 

∂θ/∂t = c (∂2θ/∂x
2
),  

Where, the parameter η= x/√ct,  giving  

the solution, 

 θ = θ0 erf (x/√4ct).  

MATHEMATICAL TREATMENT 

     Consider the two dimensional transient 

equation, 

  
∂θ

∂t
 = c  

∂2θ

∂x 2
+  

∂2θ

∂y 2
                    (1) 

Iraqi Journal of Chemical and 

Petroleum Engineering 

 

University of Baghdad 

College of Engineering 



 

 

 will be solved for an initial value of the 

function θ equal to zero, i.e. 

 

θ (0,x,y) = 0                                          (2)    

   

     Introducing the combined variable, 

  η =  
x+y

√ct
                                      (3) 

 Differentiating η with respect to t, x and y 

to find the equivalent forms of  
∂θ

∂t
 , 

∂2θ

∂x 2
  

and  
∂2θ

∂y 2
  in terms of η, thus: 

d η

dt
=  −

1

2
 
x +y

√ct
= −η/2t           (4) 

    ∴   
∂θ

∂t
=

dθ

dη
.

dη

dt
 = − 

η

2t
 
dθ

dη
        (5) 

     And  
dη

dx
=  

1

√ct
  ;   

dη

dy
=  

1

√ct
        (6) 

Therefore, 

∂2θ

∂x2
=  

∂

∂x
  
∂θ

∂x
 =  

d

dη
 .

dη

dx
  

dθ

dη
.
dη

dx
  

= 
1

ct
 

d 2θ

dη 2
                                  (7)  

Similarly, 

∂2θ

∂y 2
  = 

1

ct
 

d 2θ

    dη 2    
                    (8) 

Substituting equations 5, 7 and 8 into 

equation 1 gives, 

d 2θ

    dη 2    
+  

𝜂

4 
 
dθ

dη
  = 0              (9) 

By putting P =
dθ

dη
 and hence 

dP

dη
 = 

d 2θ

    dη 2    
 

This yields, 

dP

dη
+ η P = 0                                 (10) 

Therefore, 

P = A exp (
−η

2

8
) = 

d θ

d η
                 (11) 

       Where A is constant of  

integration. Integrating equation (11) 

gives, 

θ = A erf  
η

√8
  +  B                 (12) 

Applying the initial conditions, 

[equation (2)] leads to, 

0 = A erf (∞) + B 

But, erf (∞) = 1 

Therefore, A = -B. Then  the solution 

becomes, 

θ = B  [1 − erf  
η

√8
  ] 

θ = B  erfc  
η

√8
                             (13) 

      The constant B may be found for 

a specified boundary condition. But, 

for convenience it is assigned a value 

of constant distribution,  θ0   . Then 
the final solution after substituting for 

η is, 

θ t, x, y =  θ0   erfc 
x +y

√8ct
              (14) 

      Similar procedure is applied to the 

three dimensional equation, 

∂θ

∂t
 = c  

∂2θ

∂x2
+  

∂2θ

∂y 2
+

∂2θ

∂z 2
          (15) 

with, 

η =
 x +y +z 

√ct
                            (16) 

to give the solution, 

θ t, x, y, z = θ0 erfc   
x+y +z

√12ct
         (17) 



Aparticular  Solution of The Two And Three Dimensional  Transient Diffusion Equations                                   

 

20 
Vol.12 No.2 (June 2011) 

CONCLUSIONS 

- The particular solution obtained, here, is 

useful for in problems of transient heat and 

mass transfer in multi-dimensions. 

- The solution may be used for problems of heat 

and  mass diffusion in a flowing fluid through a 

conduit, described by the equation, 

νz   
∂θ

∂t
= c  

∂2θ

∂x2
+  

∂2θ

∂y 2
  

Where νz  is the average velocity along the 
length of the conduit. 

- The solution is also useful for statistical 

equations and in problems of stochastic nature 

such as Brownian motion. 

-  In applying the suggested solution the 

dimensions x, y and z should be 

normalized to vary between zero and 

unity and where the second derivative 

exist. 

 

NOMENCLATURE 

A, B:  Constant of integration. 

c:  Diffusion parameter. 

erf :  Error function. 

erfc: Complimentary error function. 

P:   
dθ

dη
 

t:  Independent variable (time). 

νz: Velocity in z-direction. 

x,y,z: Independent variables (linear 

dimensions). 

η: Combined variable, defined by eq. 3. 

θ: Dependant variable. 

 

 

REFERENCES 

1- Wylie, C.R. and L.C. Barrett, (1995),      

"Advanced Engineering Mathematics", 

6
th

      Edition, McGraw-Hill. 

2- Jenson, V.G. and G.V. Jeffreys, (2001), 
"Mathematical Methods in Chemical 

Engineering", 4
th

 Edition, Academic 

Press. 

3- Bird, R.B.; W.E. Stewart and E.N. 
Lightfoot, (2002), "Transport 

Phenomena", 2
nd

 Edition, John Wiley. 

4- Kreyszig, E., (1998), "Advanced 
Engineering Mathematics", 8

th
 

Edition, John Wiley. 

5- Dennis, G.Z. and M.R. Cullen, (1999), 
"Advanced Engineering 

Mathematics", 3
rd

 Edition, Jones and 

Bartlett